Chapter 10, Problem 10.41P

(a)

Interpretation Introduction

Interpretation:

The net ionic equation for the double replacement reaction of ${\text{Na}}_{\text{2}}{\text{S}}_{\left(\text{aq}\right)}$ and ${\text{HCl}}_{\left(\text{aq}\right)}$ has to be written.

Concept Introduction:

Net ionic equation:  A chemical equation, which lists only those species taking part in the reaction is called as net ionic equation.  The net ionic equation is used in acid-base neutralization reactions, double replacement reactions and redox reactions.

Spectator ions:  A spectator ion is one, which exists in the same form on both the reactant and product sides of a chemical reaction.

Answer to Problem 10.41P

The net ionic equation is,

  ${\text{2H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+S}}^{\text{2-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{H}}_{\text{2}}{\text{S}}_{\left(\text{g}\right)}$

Explanation of Solution

Solutions of sodium sulphide and hydrochloric acid react to form sodium chloride and hydrogen sulphide.  This is a double replacement reaction.  The complete equation is,

  ${\text{Na}}_{\text{2}}{\text{S}}_{\left(\text{aq}\right)}{\text{+2HCl}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{2NaCl}}_{\left(\text{aq}\right)}{\text{+H}}_{\text{2}}{\text{S}}_{\left(\text{g}\right)}$

The complete ionic equation is,

  ${\text{2Na}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+S}}^{\text{2-}}{}_{\left(\text{aq}\right)}{\text{+2H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{2Na}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+H}}_{\text{2}}{\text{S}}_{\left(\text{g}\right)}$

The spectator ions, ${\text{Na}}^{\text{+}}{}_{\left(\text{aq}\right)}$ and ${\text{Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}$ appear on the both sides of the equation.  Hydrogen sulphide does not break into ions in this equation.  Because they appear on both sides of the complete ionic equation, the spectator ions can be cancelled.

The net ionic equation is written as,

  ${\text{2H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+S}}^{\text{2-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{H}}_{\text{2}}{\text{S}}_{\left(\text{g}\right)}$

(b)

Interpretation Introduction

Interpretation:

The net ionic equation for the double replacement reaction of ${\text{PbCl}}_2{}_{\left(\text{aq}\right)}$ and ${\text{Na}}_{\text{2}}{\text{S}}_{\left(\text{aq}\right)}$ has to be written.

Concept Introduction:

Refer to part (a).

Answer to Problem 10.41P

The net ionic equation is,

  ${\text{Pb}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+S}}^{\text{2-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{PbS}}_{\left(\text{s}\right)}$

Explanation of Solution

Solutions of sodium sulphide and lead chloride react to form sodium chloride and lead sulphide.  This is a double replacement reaction.  The complete equation is,

  ${\text{PbCl}}_{\text{2}}{}_{\left(\text{aq}\right)}{\text{+Na}}_{\text{2}}{\text{S}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{2NaCl}}_{\left(\text{aq}\right)}{\text{+PbS}}_{\left(\text{s}\right)}$

The complete ionic equation is,

  ${\text{2Na}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+S}}^{\text{2-}}{}_{\left(\text{aq}\right)}{\text{+Pb}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+2Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{2Na}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+PbS}}_{\left(\text{s}\right)}$

The spectator ions, ${\text{Na}}^{\text{+}}{}_{\left(\text{aq}\right)}$ and ${\text{Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}$ appear on the both sides of the equation.  Lead sulphide does not break into ions in this equation.  Because the spectator ions appear on both sides of the complete ionic equation, the spectator ions can be cancelled.

The net ionic equation is written as,

  ${\text{Pb}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{+S}}^{\text{2-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{PbS}}_{\left(\text{s}\right)}$

(c)

Interpretation Introduction

The net ionic equation for the neutralization reaction of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}$ and ${\text{KOH}}_{\left(\text{aq}\right)}$ has to be written.

Concept Introduction:

Refer to part (a).

Answer to Problem 10.41P

The net ionic equation is,

  $\begin{array}{l}{\text{2H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2OH}}^{\text{-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\\ {\text{H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+OH}}^{\text{-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\end{array}$

Explanation of Solution

Solutions of sulphuric acid and potassium hydroxide react to form potassium sulphate and water.  This is an acid-base reaction.  The complete equation is,

  ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\left(\text{aq}\right)}{\text{+2KOH}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}{}_{\left(\text{aq}\right)}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

The complete ionic equation is,

  ${\text{2H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+SO}}_{\text{4}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}{\text{+2K}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2OH}}^{\text{-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{2K}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+SO}}_{\text{4}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

The spectator ions, ${\text{K}}^{\text{+}}{}_{\left(\text{aq}\right)}$ and ${\text{SO}}_{\text{4}}{{}^{\text{2-}}}_{\left(\text{aq}\right)}$ appear on the both sides of the equation.  Because spectator ions appear on both sides of the complete ionic equation, the spectator ions can be cancelled.

The net ionic equation is written as,

  ${\text{2H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2OH}}^{\text{-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

The above net ionic equation can be reduced as,

  ${\text{H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+OH}}^{\text{-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

(d)

Interpretation Introduction

The net ionic equation for the reaction of ${\text{Na}}_{\text{2}}{\text{O}}_{\left(\text{s}\right)}$ and ${\text{HCl}}_{\left(\text{aq}\right)}$ has to be written.

Concept Introduction:

Refer to part (a).

Answer to Problem 10.41P

The net ionic equation is,

  ${\text{2H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+O}}^{\text{2-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

Explanation of Solution

Solutions of sodium oxide and hydrochloric acid react to form sodium chloride and hydrogen sulphide.  The complete equation is,

  ${\text{Na}}_{\text{2}}{\text{O}}_{\left(\text{s}\right)}{\text{+2HCl}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{2NaCl}}_{\left(\text{aq}\right)}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

The complete ionic equation is,

  ${\text{2Na}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+O}}^{\text{2-}}{}_{\left(\text{aq}\right)}{\text{+2H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{2Na}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+2Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

The spectator ions, ${\text{Cl}}^{\text{-}}{}_{\left(\text{aq}\right)}$ and sodium appear on the both sides of the equation.  Because spectator ions on both sides of the complete ionic equation, the spectator ions can be cancelled.

The net ionic equation is written as,

  ${\text{2H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+O}}^{\text{2-}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{+H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

(e)

Interpretation Introduction

The net ionic equation for the neutralization reaction of ${\text{NH}}_{\text{3}}{}_{\left(\text{aq}\right)}$ and ${\text{HCl}}_{\left(\text{aq}\right)}$ has to be written.

Concept Introduction:

Refer to part (a).

Answer to Problem 10.41P

The net ionic equation is,

  ${\text{NH}}_{\text{3(aq)}}{\text{+H}}^{\text{+}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{NH}}_{\text{4}}{{}^{\text{+}}}_{\left(\text{aq}\right)}$

Explanation of Solution

Solutions of ammonia and hydrochloric acid react to form ammonium chloride.  This is a neutralization reaction.  The complete equation is,

  ${\text{NH}}_{\text{3}}{}_{\left(\text{aq}\right)}{\text{+HCl}}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{NH}}_{\text{4}}{\text{Cl}}_{\left(\text{aq}\right)}$

The complete ionic equation is,

  ${\text{NH}}_{\text{3(aq)}}{\text{+H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+Cl}}^{\text{-}}{}_{\text{(aq)}}\stackrel{}{\to }{\text{NH}}_{\text{4}}{{}^{\text{+}}}_{\left(\text{aq}\right)}{\text{+Cl}}^{\text{-}}{}_{\text{(aq)}}$

The net ionic equation is written as,

  ${\text{NH}}_{\text{3(aq)}}{\text{+H}}^{\text{+}}{}_{\left(\text{aq}\right)}\stackrel{}{\to }{\text{NH}}_{\text{4}}{{}^{\text{+}}}_{\left(\text{aq}\right)}$